A couple of GRE questions

Question

Look for help with the following GRE questions

Question 1. If $C$ is the circle in the complex plane whose equation $|z|=\pi$, oriented counterclockwise, find the value of the integral $\oint_C(\cos z-z\cos\frac{1}{z})dz$.

Question 2. How to show the sequence $\{x_n\}_{n=1}^\infty$ definted by $x_{n+1}=\frac{1}{2}(x_n+\frac{2}{x_n})$, $x_1\ne 0$ converge.

Question 3. Let $L$ be the curve whose equation in the polar coordinates $r$ and $\theta$ is $r^2=4\cos 2\theta$. Fine the largest value of $y$ such that the point with rectangular coordinates $(x,y)$ is on $L$.

Question 4. Consider the set $S$ of all real-valued functions defined on $[a, b]$, $a<b$. Is it true that if the inverse of $f$ is a constant function, then $f$ is a constant function?

Answer 1

For the first, $\oint_C \cos z dz = 0$ since $\cos$ is an entire function. Thus, only the term $z\cos\frac{1}{z}$ is relevant. That term has a single pole at $z=0$, and by the residue theorem you therefore have $$ \oint_C z\cos\frac{1}{z} dz = i2\pi R $$ where $R$ is the residue (i.e. the coefficient of $z^{-1}$ in the laurent expansion of $z\cos\frac{1}{z}$) at $0$. Now, $\cos z = \frac{1}{2}(e^{iz} + e^{-iz})$, hence the coefficient of $z^2$ in the taylor expansion of $\cos z$ is $\frac{1}{2}(\frac{-1}{2} + \frac{-1}{2})$ = $-\frac{1}{2}$. Which is the same as the coefficient of $z^{-2}$ in the laurent expansion of $\cos\frac{1}{z}$, and hence the coefficient of $z^{-1}$ in the laurent expansion of $z\cos\frac{1}{z}$, i.e. the residue. Together, you get that $$ \oint_C \cos z - z\cos\frac{1}{z} dz = 0 - i2\pi\frac{-1}{2} = i\pi $$

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For the second, first find a fixed point, i.e. solve $x_{n+1} = x_n$. You get $$ \frac{1}{2}\left(x_n + \frac{2}{x_n}\right) = x_n \iff \frac{1}{x_n} = \frac{x_n}{2} \iff x_n = \sqrt{2} $$ Then show that $\sqrt{2} < x_{n+1} < x_n$ if $x_n > \sqrt{2}$, which works similarly $$ \begin{eqnarray} \sqrt{2} < \frac{1}{2}\left(x_n + \frac{2}{x_n}\right) < x_n &\iff& \sqrt{2} - \frac{x_n}{2} < \frac{1}{x_n} < \frac{x_n}{2} \iff \\ 2\sqrt{2}x_n - x_n^2 < 2 < x_n^2 &\iff& 2 - (x_n - \sqrt{2})^2 < 2 < x_n^2 \end{eqnarray} $$ The last inequality shows not only that the sequence decrases monotonically if started at any value larger than $\sqrt{2}$, it also shows that regardless of where it it started, it always immediatly assumes a value larger than $\sqrt{2}$. Thus, the sequence converges regardless of its initial value (Strictly speaking, the above verified that only for positive $x_1$, but things work identically for negative $x_1$)

Answer 2

For $1$, what I'd do is ignore the $\cos z$ term since $\cos$ is an entire function so its integrals around closed curves vanish; then write the other term's Laurent series $-z(1-1/2z^2+...)=-z+1/2z-...$ to get that the residue at $0$ is $1/2$. Clearly there are no other poles.

For question $2$, note that on the GRE it's not important to be able to show very rigorously that the series converges; but note that it increases for $x_n<\sqrt{2}$ and decreases for $x_n>\sqrt{2}$ (this is taking $x_1>0$ and leaving the negative case to you) so it at least won't diverge to infinity. Then to find the limit $L$ assuming it exists, in this sort of problem one takes the limit of both sides and writes $L=1/2(L+2/L)$, then solves the resulting quadratic. As I've already hinted, the positive solution is $\sqrt{2}$.

I can't think of a very good way to do question $3$ quickly. Rewriting into rectangular coordinates and doing a grotesque implicit differentiation would take rather more than the $3$ minutes I aim for on GRE problems. But it's multiple choice, and we can at least see that we get up to $y=1/\sqrt{2}$ at $\theta=\pi/6$ and by $\theta=\pi/4$ we're down to $0$.

As to the final question, remember that only one-to-one functions can have inverses, and that the inverses are also one-to-one. That is, there is no such $f$.